Euler-Maclaurin求和公式估计梯形积分公式的误差

西西在大学群里的一道题,也是2014年第六届非数竞赛预赛最后一题的推广:

设${A_n} = \frac{n}{{{n^2} + 1}} + \frac{n}{{{n^2} + {2^2}}} + \cdots + \frac{n}{{{n^2} + {n^2}}}$,求极限

\[\mathop {\lim }\limits_{n \to \infty } {n^4}\left( {\frac{1}{{24}} - n\left( {n\left( {\frac{\pi }{4} - {A_n}} \right) - \frac{1}{4}} \right)} \right).\]

这里提供个一般的方法.

Euler-Maclaurin求和公式

设函数$f\in C^{(2m+2)}[a,b],h=\frac{b-a}{n},x_i=a+ih,i=0,1,\cdots,n$,则

\begin{align*}\frac{{b - a}}{n}\sum\limits_{i = 1}^n {\frac{1}{2}\left[ {f\left( {{x_{i - 1}}} \right) + f\left( {{x_i}} \right)} \right]} - \int_a^b {f\left( x \right)dx} = &\sum\limits_{k = 1}^m {\frac{{{B_{2k}}}}{{\left( {2k} \right)!}}{h^{2k}}\left[ {{f^{\left( {2k - 1} \right)}}\left( b \right) - {f^{\left( {2k - 1} \right)}}\left( a \right)} \right]} \\&+ \frac{{{B_{2m + 2}}}}{{\left( {2m + 2} \right)!}}{h^{2m + 2}}{f^{\left( {2m + 2} \right)}}\left( \xi \right)\left( {b - a} \right),\end{align*}

其中$\xi\in [a,b]$, $B_{2k}(k=1,2,\cdots,m+1)$是Bernoulli数且$B_2=\frac16,B_4=-\frac{1}{30},B_6=\frac{1}{42}$.

解:取$a=0,b=1,f(x)=\frac{1}{1+x^2}$,则$h=\frac1n,x_i=\frac{i}{n},A_n=\frac{1}{n}\sum\limits_{i = 1}^n {f\left( {{x_i}} \right)}$,则

\begin{align*}&{A_n} + \frac{1}{{4n}} - \frac{\pi }{4} = \frac{1}{2}\left[ {\left( {{A_n} - \frac{1}{{2n}} + \frac{1}{n}} \right) + {A_n}} \right] - \frac{\pi }{4} = \frac{{{B_2}}}{{2!}} \cdot \frac{1}{{{n^2}}}\left[ {f'\left( 1 \right) - f'\left( 0 \right)} \right]\\+ &\frac{{{B_4}}}{{4!}} \cdot \frac{1}{{{n^4}}}\left[ {f'''\left( 1 \right) - f'''\left( 0 \right)} \right] + \frac{{{B_6}}}{{6!}} \cdot \frac{1}{{{n^6}}}\left[ {{f^{\left( 5 \right)}}\left( 1 \right) - {f^{\left( 5 \right)}}\left( 0 \right)} \right] + \frac{{{B_8}}}{{8!}} \cdot \frac{1}{{{n^8}}}{f^{\left( 8 \right)}}\left( \xi \right),\end{align*}

其中$\xi\in [0,1]$,也即

\[{n^4}\left( {\frac{1}{{24}} - n\left( {n\left( {\frac{\pi }{4} - {A_n}} \right) - \frac{1}{4}} \right)} \right) = \frac{1}{{2016}} + \frac{{{B_8}}}{{8!}} \cdot \frac{1}{{{n^2}}}{f^{\left( 8 \right)}}\left( \xi \right),\]

注意到${f^{\left( 8 \right)}}\left( \xi \right)$有界,因此$n\to\infty$时,所求极限为$\frac{1}{{2016}}$.